YES

Problem 1: 

(VAR v_NonEmpty:S x:S)
(RULES
f(0) -> 1
f(s(x:S)) -> g(f(x:S))
f(s(x:S)) -> +(f(x:S),s(f(x:S)))
g(x:S) -> +(x:S,s(x:S))
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(s(x:S)) -> F(x:S)
 F(s(x:S)) -> G(f(x:S))
-> Rules:
 f(0) -> 1
 f(s(x:S)) -> g(f(x:S))
 f(s(x:S)) -> +(f(x:S),s(f(x:S)))
 g(x:S) -> +(x:S,s(x:S))

Problem 1: 

SCC Processor:
-> Pairs:
 F(s(x:S)) -> F(x:S)
 F(s(x:S)) -> G(f(x:S))
-> Rules:
 f(0) -> 1
 f(s(x:S)) -> g(f(x:S))
 f(s(x:S)) -> +(f(x:S),s(f(x:S)))
 g(x:S) -> +(x:S,s(x:S))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(s(x:S)) -> F(x:S)
->->-> Rules:
 f(0) -> 1
 f(s(x:S)) -> g(f(x:S))
 f(s(x:S)) -> +(f(x:S),s(f(x:S)))
 g(x:S) -> +(x:S,s(x:S))

Problem 1: 

Subterm Processor:
-> Pairs:
 F(s(x:S)) -> F(x:S)
-> Rules:
 f(0) -> 1
 f(s(x:S)) -> g(f(x:S))
 f(s(x:S)) -> +(f(x:S),s(f(x:S)))
 g(x:S) -> +(x:S,s(x:S))
->Projection:
 pi(F) = 1

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 f(0) -> 1
 f(s(x:S)) -> g(f(x:S))
 f(s(x:S)) -> +(f(x:S),s(f(x:S)))
 g(x:S) -> +(x:S,s(x:S))
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.